Anonymous
Anonymous asked in Science & MathematicsMathematics · 4 weeks ago

Help needed with a probability question, please?

Four teams Antwerp, Brindisi, Cadiz and Dortmund are left in the championship. One (and only one) of the teams can win the trophy. Antwerp is twice as likely to win as Brindisi, Brindisi is three times as likely to win as Cadiz and Cadiz are twice as likely to win as Dortmund.

What are the teams respective probabilities of winning?

What is the probability that neither Brindisi nor Cadiz wins?

1 Answer

Relevance
  • 4 weeks ago
    Favourite answer

    Let a, b, c and d be the probabilities that Antwerp, Brindisi, Cadiz and Dortmund win the trophy respectively.

    a + b + c + d = 1 …. [1]

    a = 2b …… [2]

    b = 3c …… [3]

    c = 2d …… [4]

    Substitute [4] into [3]:

    b = 3(2d)

    b = 6d …… [5]

    Substitute [5] into [1]:

    a = 2(6d)

    a = 12d …… [6]

    Substitute [4], [5] and [6] into [1]:

    12d + 6d + 2d + d = 1

    21d = 1

    d = 1/21

    P(Antwerp wins) = 12 × (1/21) = 4/7

    P(Brindisi wins) = 6 × (1/21) = 2/7

    P(Cadiz wins) = 2 × (1/21) = 2/21

    P(Dortmund wins) = 1/21

    P(neither Brindisi nor Cadiz wins)

    = P(either Antwerp or Dortmund wins)

    = P(Antwerp wins) + P(Dortmund wins)

    = (4/7) + (1/21)

    = (12/21) + (1/21)

    = 13/21

    • Commenter avatarLog in to reply to the answers
Still have questions? Get answers by asking now.